OFFSET
0,2
COMMENTS
Sequence appears in A096952 (upper bounds for Lagrange remainder in Taylor expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2)).
Divisibility of 2^(2*n+1) + 3^(2*n+1) by 5 is proved by induction.
The sequence a(n+1), with g.f. (7-36*x)/(1-13*x+36*x^2) and formula (27*9^n + 8*4^n)/5, is the Hankel transform of C(n) + 6*C(n+1), where C(n) is A000108(n). - Paul Barry, Dec 06 2006
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,-36).
FORMULA
a(n) = (2^(2*n+1) + 3^(2*n+1))/5.
G.f.: (1-6*x)/((1-4*x)*(1-9*x)).
From Reinhard Zumkeller, Mar 07 2008: (Start)
a(n+1) = 4*a(n) + 3^(2*n+1), a(0) = 1.
a(n) = A138233(n)/5. (End)
From Elmo R. Oliveira, Aug 02 2025: (Start)
E.g.f.: exp(4*x)*(2 + 3*exp(5*x))/5.
a(n) = 13*a(n-1) - 36*a(n-2).
a(n) = A015441(2*n+1). (End)
MATHEMATICA
LinearRecurrence[{13, -36}, {1, 7}, 19] (* Ray Chandler, Jul 14 2017 *)
PROG
(Magma) [(2^(2*n+1) + 3^(2*n+1))/5: n in [0..30]]; // Vincenzo Librandi, May 31 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jul 16 2004
STATUS
approved